Question

Solve \(\left(x^{2}-y^{2}\right) d x+2 x y d y=0, y(1)=2\)

Short answer

Question: Solve the given first-order homogeneous differential equation and find the function y(x) that satisfies the initial condition y(1) = 2: \(\frac{dy}{dx} + \frac{2xy}{x^2-y^2} = 0\) Answer: The solution of the given differential equation in implicit form is: \(\frac{1}{2}\left(\frac{y^2}{x^2}\right)\ln\left|\frac{y^2}{x^2}\right| - \frac{1}{2} \frac{y^2}{x^2} = -\ln|x| + 4\ln2 - 2\) with the initial condition y(1) = 2.

Step-by-step solution

Rewrite the equation

Notice that the given equation can be rewritten as: \(\frac{dy}{dx} + \frac{2xy}{x^2-y^2} = 0\)

Separate the variables and integrate

Divide by \(2xy\) on both sides and multiply with \(dx\) to separate the variables: \(\frac{dy}{y} = -\frac{dx}{x}\cdot\frac{x^2}{x^2 - y^2}\) Now, integrate both sides with respect to their corresponding variables: \(\int \frac{1}{y} dy = -\int \frac{x^2}{x^2 - y^2} dx\) On the left side, you get: \(\ln |y| = -\int \frac{x^2}{x^2 - y^2} dx + C_1\)

Solve for y(x)

To solve for \(y(x)\), we are going to use substitution. Let \(y = xu\), then \(dy = xdu + udx\). Substitute this into the equation and express everything in terms of \(u\) and \(x\): \(\int \frac{xdu + udx}{xu} = -\int \frac{x^2}{x^2 - x^2u^2} dx\) Now, divide both sides by x: \(\int \frac{du + u \frac{dx}{x}}{u} = -\int \frac{1}{1-u^2} dx\) To solve the integral on the left, we can differentiate \(v=u^2\) to get \(dv=2udu\). Now the left side can be rewritten as \(\int \frac{1}{2} dv + \frac{1}{2} v \frac{dx}{x} = -\int \frac{1}{1-v} dx\) Integrating both sides yields \(\frac{1}{2}v\ln|v| - \frac{1}{2} v = -\ln|x| + C\) We remember that \(v = u^2\), and thus we obtain \(\frac{1}{2}u^2\ln|u^2| - \frac{1}{2} u^2 = -\ln|x| + C\)

Substitute back for y(x) and impose initial condition

Now that we have our expression in terms of \(u\), we can substitute back for \(y(x)\), using \(u = y/x\): \(\frac{1}{2}\left(\frac{y^2}{x^2}\right)\ln\left|\frac{y^2}{x^2}\right| - \frac{1}{2} \frac{y^2}{x^2} = -\ln|x| + C\) Now, apply the initial condition \(y(1) = 2\): \(\frac{1}{2}\left(\frac{2^2}{1^2}\right)\ln\left|\frac{2^2}{1^2}\right| - \frac{1}{2} \frac{2^2}{1^2} = -\ln|1| + C\) Which yields \(C = 4\ln2 - 2\).

Write the final solution

Our final solution in implicit form, with the calculated constant, is: \(\frac{1}{2}\left(\frac{y^2}{x^2}\right)\ln\left|\frac{y^2}{x^2}\right| - \frac{1}{2} \frac{y^2}{x^2} = -\ln|x| + 4\ln2 - 2\)

Key concepts

Separable Differential Equations

Separable differential equations are a subclass of ordinary differential equations (ODEs) where the variables can be separated on opposite sides of the equation. This separation simplifies the integration process. Consider an ODE in the form \(\frac{dy}{dx} = g(x)h(y)\), where \(g(x)\) and \(h(y)\) are functions of \(x\) and \(y\), respectively. The goal is to rearrange the equation so that each variable and its differential are on the same side, i.e., \(\frac{1}{h(y)}dy = g(x)dx\).

Once this is achieved, you integrate both sides independently. If \(H(y)\) is an antiderivative of \(1/h(y)\) and \(G(x)\) is an antiderivative of \(g(x)\), then post-integration, the general solution is expressed as \(H(y) = G(x) + C\), where \(C\) is the constant of integration. In the given exercise, the strategy to rewrite the equation and separate the variables was the first essential step toward finding the solution.

Integration Techniques

Successful integration relies on recognizing which integration technique to apply to a given function. The main techniques include substitution, integration by parts, partial fractions, and trigonometric substitutions. In the provided solution, substitution is used to simplify the integral.

More specifically, when faced with an integral that cannot be easily computed, such as \(\int \frac{x^2}{x^2 - y^2} dx\), we can simplify the process by substituting \(u\) for \(y/x\), reducing the integral to a more manageable form. Another example is integrating \(\int \frac{dv}{1-v} dx\) after substitution, which is a standard form easily computed. The substitution technique is versatile and powerful in making seemingly complex integrals solvable.

Initial Value Problem

An initial value problem (IVP) in the context of differential equations is one where the solution is required to satisfy an additional condition at a specific point, typically where the value of the function is given. The IVP ensures that the solution to the differential equation is unique. Once the general solution is found through integration, the initial condition provides an exact value for the constant of integration \(C\).

In our exercise, the IVP is given by \(y(1)=2\), which means that when \(x=1\), \(y\) must be 2. By substituting these values into the general solution, the value of \(C\) is determined, allowing us to write the particular solution to the differential equation that conforms to the initial condition.

Implicit Differentiation

Implicit differentiation is used when dealing with equations where \(y\) cannot be explicitly solved in terms of \(x\), or it is simply cumbersome to do so. This technique involves differentiating both sides of an equation with respect to \(x\), treating \(y\) as an implicit function of \(x\). When doing so, the derivative of \(y\) with respect to \(x\), denoted as \(\frac{dy}{dx}\), will appear naturally in the process.

This approach is particularly useful when we want to find the rate at which \(y\) is changing with respect to \(x\) without explicitly solving for \(y\) as a function of \(x\). In the context of the exercises, implicit differentiation was a crucial step in the derivation of the final expression that conforms with the given initial condition.